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\(a,x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
\(b,25-4x^2-4xy-y^2\)
\(=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x+y\right)\)
\(c,x^3-x+y^3-y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+1\right)\)
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
1, \(7x^2y^5-14x^3y^4-21y^3=7y^3\left(x^2y^2-2x^3y-3\right)\)
2, \(-12x^2y+6xy^2-15xy=3xy\left(-4x+2y-5\right)\)
3, \(-14x^2y-21xy^2+28x^2y^2=-7xy\left(2x+3y-4xy\right)\)
\(a,x^2+7x+7y-y^2\)
\(=x^2-y^2+7\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
\(b,x^2-2x-9y^2+6y\)
\(=x^2-\left(3y\right)^2-2\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-2\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-2\right)\)
\(c,x^2-xy+x^3-3x^{2y}+3x^{2y}-y^3\)
\(=x\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x+x^2+xy+y^2\right)\)
\(1,\)
\(\left(x^2-9y^2\right)\left(4x+12y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-4\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x-3y-4\right)\)
\(3,\)
\(-x^2+2xy-y^2+25\)
\(=-\left(x^2-2xy+y^2\right)+25\)
\(=25-\left(x-y\right)^2\)
\(=5^2-\left(x-y\right)^2\)
\(=\left(5-x+y\right)\left(5+x-y\right)\)
f) = x2( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )
g) = 4( x - y ) + ( x - y )2 = ( x - y )( x - y + 4 )
h) = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )
j) = ( x - y )( x2 + xy + y2 ) - 3( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
Trả lời:
f, x3 - 4x2 - 9x + 36 = ( x3 - 4x2 ) - ( 9x - 36 ) = x2 ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x2 - 9 ) = ( x - 4 )( x - 3 )( x + 3 )
g, 4x - 4y + x2 - 2xy + y2 = ( 4x - 4y ) + ( x2 - 2xy + y2 ) = 4 ( x - y ) + ( x - y )2 = ( x - y ) ( 4 + x - y )
h, x4 + x3 + x2 - 1 = ( x4 + x3 ) + ( x2 - 1 ) = x3 ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i, x2 - y2 - 4x - 4y = ( x2 - y2 ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )
j, x3 - y3 - 3x + 3y = ( x3 - y3 ) - ( 3x - 3y ) = ( x - y )( x2 + xy + y2 ) - 3 ( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
\(\frac{x+y}{\left(x+3\right)^2+\left(y-2\right)^2}\)
ĐKXĐ : \(\left(x+3\right)^2+\left(y-2\right)^2\ne0\)
⇔ \(\hept{\begin{cases}\left(x+3\right)^2\ne0\\\left(y-2\right)^2\ne0\end{cases}}\)
⇔ \(\hept{\begin{cases}x+3\ne0\\y-2\ne0\end{cases}}\)
⇔ \(\hept{\begin{cases}x\ne-3\\y\ne2\end{cases}}\)